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php - 备份mysql数据库取特定表

问题描述

你好这是我用来备份mysql数据库的php代码，它需要所有的数据库和所有的表，我想在数据库中获取一个特定的表而不是所有的表。可能有人请告诉我如何修改 php 代码以获取单个表而不是所有表？这是代码：

<?php

/**
* Updated: Mohammad M. AlBanna
* Website: MBanna.info
*/


//MySQL server and database
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'novtech';
$tables = '*';

//Call the core function
backup_tables($dbhost, $dbuser, $dbpass, $dbname, $tables);

//Core function
function backup_tables($host, $user, $pass, $dbname, $tables = '*') {
    $link = mysqli_connect($host,$user,$pass, $dbname);

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
        exit;
    }

    mysqli_query($link, "SET NAMES 'utf8'");

    //get all of the tables
    if($tables == '*')
    {
        $tables = array();
        $result = mysqli_query($link, 'SHOW TABLES');
        while($row = mysqli_fetch_row($result))
        {
            $tables[] = $row[0];
        }
    }
    else
    {
        $tables = is_array($tables) ? $tables : explode(',',$tables);
    }

    $return = '';
    //cycle through
    foreach($tables as $table)
    {
        $result = mysqli_query($link, 'SELECT * FROM '.$table);
        $num_fields = mysqli_num_fields($result);
        $num_rows = mysqli_num_rows($result);

        $return.= 'DROP TABLE IF EXISTS '.$table.';';
        $row2 = mysqli_fetch_row(mysqli_query($link, 'SHOW CREATE TABLE '.$table));
        $return.= "\n\n".$row2[1].";\n\n";
        $counter = 1;

        //Over tables
        for ($i = 0; $i < $num_fields; $i++) 
        {   //Over rows
            while($row = mysqli_fetch_row($result))
            {   
                if($counter == 1){
                    $return.= 'INSERT INTO '.$table.' VALUES(';
                } else{
                    $return.= '(';
                }

                //Over fields
                for($j=0; $j<$num_fields; $j++) 
                {
                    $row[$j] = addslashes($row[$j]);
                    $row[$j] = str_replace("\n","\\n",$row[$j]);
                    if (isset($row[$j])) { $return.= '"'.$row[$j].'"' ; } else { $return.= '""'; }
                    if ($j<($num_fields-1)) { $return.= ','; }
                }

                if($num_rows == $counter){
                    $return.= ");\n";
                } else{
                    $return.= "),\n";
                }
                ++$counter;
            }
        }
        $return.="\n\n\n";
    }

    //save file , db-backup
    //$fileName = 'novtechDB-'.time().'-'.(md5(implode(',',$tables))).'.sql';
    $fileName = 'novtechDB'.'.sql';
    $handle = fopen($fileName,'w+');
    fwrite($handle,$return);

   if(fclose($handle)){
        echo "Done, the file name is: ".$fileName;
        exit; 
    }
}

我把 $tables= array('myTable') 放在表格 array[foreach($tables as $table)] 的循环之前，我现在可以工作，但它给了我一个错误：注意：未定义的变量：返回。请问有人可以帮我找出问题所在吗？如何解决？

标签： phpmysqldatabasebackup

解决方案

在表格数组的循环之前，将数组设置为你要抓取的表格。

$tables = ['tale1','table2','table3'];
foreach($tables as $table)

删除被注释的代码块//get all of the tables

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